Test: Limits And Derivatives (CBSE Level) - 1

We have:
limx→0 ln(1+ax)]/x=a
lim x→0 ln(1+ax)/ax=a
and then:
lim x→0 [ln(1+ax)-ln(1+bx)]/x
= a-b

On rationalizing,
 Lt x-->0 (1-Cosx)/[√(1+x) +1]
= lt x-->0 2sin²(x/2)[√(1+x)+1] /x
= lt x-->0 [2sin(x/2)/x].sinx/2 [√(1+x) +1 ]
= 2.1/2 .0[1+1]=0

f(0) = lim(x→0) x sin (1/x)
​We know ∀ x ∈ R,sin (1/x) ∈ [−1,1]
Hence, f(0) = lim(x→0) xsin(1/x) 
= 0

lim(x→0) (tanx−sinx)/x³
=lim(x→0) [sinx(x−cosx)]/x³cosx
=lim(x→0) (sinx)/x × (1−cosx)/x² × (1/cosx)
= 1 × 1/2 × 1/1
= 1/2

lim x→0 (1−cosx)/(xlog(1+x))
= lim x→0 {2sin² x/2}/[xlog(1+x)]
= lim x→0 {{2sin²x/2/(x/2)²}*/1/4}/[1/xlog(1+x)]
= lim x→0 [log(1+x)/x]1/2
= 2​

Consider the given function.
lim x→0 ((1-x)ⁿ −1)/x)
This is 0/0 form.
So, apply L-Hospital rule,
lim x→0 [n(1-x)ⁿ−1 −0)]/1
lim x→ 0(n(1-x)ⁿ−1)
=n(1-0)ⁿ−1
= n - 1

lim(x → 0) (tanx-x)/x² tanx
As we know that tan x = sinx/cosx
lim(x → 0) (sinx/cosx - x)/x²(sinx/cosx)
lim(x → 0) (sinx - xcosx)/(x² sinx)
lim(x → 0) cosx - (-xsinx + cosx)/(x²cosx + sinx (2x))
lim(x → 0) (cosx + xsinx - cosx)/x²cosx + 2xsinx)
lim(x → 0) sinx/(xcosx + 2sinx)
Hence it is 0/0 form, apply L hospital rule
lim(x → 0) cosx/(-xsinx + cosx + 2cosx)
⇒ 1/(0+1+2)
= 1/3

 lim x-->0 ((x^2cosx-)/1-cosx)
Differentiating numerator and denominator
lim x-->0 (2xcosx - x^2sinx)/sinx
Differentiating it for a second time
lim x-->0 (2cos - 2xsinx - 2xsinx - x^2cosx)/cosx
Substituting x = 0
= lim x-->0  2/1
= 2

Suppose that the Reqd. Limit L = lim x→0 (sinx − x)/x³.
Substiture x = 3y , so that, as x→0 , y→0.
∴ L = lim y→0 sin3y − 3y/(3y)³,
= lim y→0 (3siny − 4sin3y)− 3y)/27y³,
= lim y→0 {3(siny − y)/27y³) − (4sin³y/27y³)},
⇒ L = lim y→0 1/9 * (siny − y)/y³) − 4/27* (siny/y)³...(∗)
.Note that, here,
= lim y→0(siny − y)/y³)
= lim x→0 (sinx−x)/x³) = L.
Therefore, (∗) ⇒ L = 1/9*L − 4/27
or,  8/9L = −4/27
Hence, L = −4/27*9/8
=−1/6

lim(h → 0) [sin²(x+h) - sin²x]/h
l = lim(h→0) = [2sin(x+h)cos(x+h)]/1
{As we know that sin2x = 2sinxcosx
= sin2x

- lim x-->k+ = x-{x}
Lim x→k+   k-{k}
                  = 0           

Correct Answer :- C

Explanation : lt(x->a-) -|x-a|/x-a

= - |-a-a|/(-a-a)

= -|-2a|/(-2a)

= -2a/(-2a)

= 1

√x / √(16+√x) - 4

= (√x * √(16+√x) + 4) / ( √(16+√x) - 4) * (√(16+√x) + 4)

= (√x * √(16+√x) + 4) / 16+√x - 16

= √(16+√x) + 4

now at lim x->0

= √16 + 4 = 8

Given, Lim_x→0 (cos x)^{cot² x}
= Lim_x→0 (1 + cos x – 1)^{cot² x}
= eLim_x→0 (cos x – 1) × cot² x
= eLim_x→0 (cos x – 1) / tan² x
= e-1/2